Answer:
Between 15.95 ounces and 16.15 ounces.
Step-by-step explanation:
We have the following value m, being the mean, sd, being the standard deviation and n, the sample size:
m = 16.05
sd = 0.1005
n = 4
We apply the formula of this case, which would be:
m + - 2 * sd / (n ^ 1/2)
In this way we create a range, replacing we have:
16.05 + 2 * 0.1005 / (4 ^ 1/2) = 16.1505
16.05 - 2 * 0.1005 / (4 ^ 1/2) = 15.9495
Which means that 95% of all samples are between 15.95 ounces and 16.15 ounces.
Answer:
54$
Step-by-step explanation:
- To get the tax times the number to the percentage
- 50 * 8% = 4
- 50 + 4 = 54 $
32/28=16/x
32x= 448
x=14
There are 14 red crayons in classroom
Answer:
1) 25x + 15y = 750
2) 25. On a graph it would look like : (25,0)
3) 15. On a graph it would look like : (15,0)
4) 25x - 750 = 15y or 15y - 750 = 25x
Step-by-step explanation:
Answer:
![P(35-2 < X 35+2) = P(33< X< 37)= P(X](https://tex.z-dn.net/?f=%20P%2835-2%20%3C%20X%2035%2B2%29%20%3D%20P%2833%3C%20X%3C%2037%29%3D%20P%28X%3C37%29%20-P%28X%3C32%29)
And using the cumulative distribution function we got:
![P(35-2 < X 35+2) = P(33< X< 37)= P(X](https://tex.z-dn.net/?f=%20P%2835-2%20%3C%20X%2035%2B2%29%20%3D%20P%2833%3C%20X%3C%2037%29%3D%20P%28X%3C37%29%20-P%28X%3C32%29%20%3D%20%5Cfrac%7B37-20%7D%7B50-20%7D%20-%5Cfrac%7B33-20%7D%7B50-20%7D%20%3D0.567-0.433%3D0.134%20)
The probability that preparation is within 2 minutes of the mean time is 0.134
Step-by-step explanation:
For this case we define the following random variable X= (minutes) for a lab assistant to prepare the equipment for a certain experiment , and the distribution for X is given by:
![X \sim Unif (a= 20, b =50)](https://tex.z-dn.net/?f=%20X%20%5Csim%20Unif%20%28a%3D%2020%2C%20b%20%3D50%29)
The cumulative distribution function is given by:
![F(x) = \frac{x-a}{b-a} , a \leq X \leq b](https://tex.z-dn.net/?f=%20F%28x%29%20%3D%20%5Cfrac%7Bx-a%7D%7Bb-a%7D%20%2C%20a%20%5Cleq%20X%20%5Cleq%20b)
The expected value is given by:
![E(X) = \frac{a+b}{2} = \frac{20+50}{2}=35](https://tex.z-dn.net/?f=%20E%28X%29%20%3D%20%5Cfrac%7Ba%2Bb%7D%7B2%7D%20%3D%20%5Cfrac%7B20%2B50%7D%7B2%7D%3D35)
And we want to find the following probability:
![P(35-2 < X 35+2) = P(33< X< 37)](https://tex.z-dn.net/?f=%20P%2835-2%20%3C%20X%2035%2B2%29%20%3D%20P%2833%3C%20X%3C%2037%29)
And we can find this probability on this way:
![P(35-2 < X 35+2) = P(33< X< 37)= P(X](https://tex.z-dn.net/?f=%20P%2835-2%20%3C%20X%2035%2B2%29%20%3D%20P%2833%3C%20X%3C%2037%29%3D%20P%28X%3C37%29%20-P%28X%3C32%29)
And using the cumulative distribution function we got:
![P(35-2 < X 35+2) = P(33< X< 37)= P(X](https://tex.z-dn.net/?f=%20P%2835-2%20%3C%20X%2035%2B2%29%20%3D%20P%2833%3C%20X%3C%2037%29%3D%20P%28X%3C37%29%20-P%28X%3C32%29%20%3D%20%5Cfrac%7B37-20%7D%7B50-20%7D%20-%5Cfrac%7B33-20%7D%7B50-20%7D%20%3D0.567-0.433%3D0.134%20)
The probability that preparation is within 2 minutes of the mean time is 0.134